# How do you find three consecutive even integers have a sum of -36?

Mar 10, 2018

See a solution process below:

#### Explanation:

Let's call the first even integer: $n$

Then the next two even integers would be: $n + 2$; $n + 4$

Sum means to add so we can write the following equation and solve for $n$:

$n + \left(n + 2\right) + \left(n + 4\right) = - 36$

$n + n + 2 + n + 4 = - 36$

$n + n + n + 2 + 4 = - 36$

$1 n + 1 n + 1 n + 6 = - 36$

$\left(1 + 1 + 1\right) n + 6 = - 36$

$3 n + 6 = - 36$

$3 n + 6 - \textcolor{red}{6} = - 36 - \textcolor{red}{6}$

$3 n + 0 = - 42$

$3 n = - 42$

$\frac{3 n}{\textcolor{red}{3}} = - \frac{42}{\textcolor{red}{3}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} n}{\cancel{\textcolor{red}{3}}} = - 14$

$n = - 14$

Then:

$n + 2 = - 14 + 2 = - 12$

$n + 4 = - 14 + 4 = - 10$

#(-14) + (-12) + (-10) = -36

The three consecutive even integers adding up to -36 are:

-14, -12, -10

Mar 10, 2018

$- 14 , - 12 , - 10.$

#### Explanation:

Take three consecutive even integers:
$\textcolor{red}{\left(2 n - 2\right) , 2 n , \left(2 n + 2\right)}$
$\therefore$ sum $= \left(2 n - 2\right) + 2 n + \left(2 n - 2\right) = - 36$
$\implies 2 n - \cancel{2} + 2 n + 2 n + \cancel{2} = - 36$
$\implies 6 n = - 36 \implies n = - 6$
So,the required integers:
$\textcolor{red}{2 n - 2} = 2 \left(- 6\right) - 2 = - 14$
$\textcolor{red}{2 n} = 2 \left(- 6\right) = - 12$
$\textcolor{red}{2 n + 2} = 2 \left(- 6\right) + 2 = - 10$